# Momentum of photon close the sun or black hole

1. Sep 11, 2006

### exponent137

We know, that photon change direction for 1.75'', when it flies close to the sun. This angle is proportional to radius of the sun. If we calculate clasically the angle is 0.82'' (arc second).

Which is momentum, which is given to sun (or black hole) by the photon change of direction. Is the same as change of momentum of the photon. Must we calculate change of momenum of the photon clasicaly or relativistically?

2. Sep 11, 2006

### pervect

Staff Emeritus
The good news is that the system of the sun + photon does have a conserved total momentum, as long as the sun and photon are essentially alone in the universe (the technical name for the requirement is that they are in an asymptotically flat space-time).

The bad news is that it's not particularly easy to calculate this momeuntm. To be really technically correct, I think you'd have to use the ADM momentum.

While it is mostly concerned with mass, http://en.wikipedia.org/wiki/Mass_in_General_Relativity talks a little bit about momentum in General relativity. Note that the article talks about how time translation symmetries generate conserved energies. The parallel argument is that space translation symetries generate conserved momenta.

The following argument is a little suspect, but I think it might work. The Komar energy of the photon is a little easier to deal with than the ADM energy - essentially, it just gets multiplied by a "red-shift" factor, which is equal to the square root of g00, the metric coefficient for time. There isn't any Komar momentum defined that I'm aware of - Komar energy is defined for static systems, and the photon isn't static - but one might guess that the momentum of the photon similarly gets multiplied by the same red-shift factor that the energy was multiplied by, i.e. the square root of g00.

If you're not familiar with momentum in special relativity, you might want to read a little bit about it here, first, before reading the next remark:
http://en.wikipedia.org/wiki/4-momentum

Basically, we expect that the length of the ADM energy-momentum 4-vector will be zero, because the invariant mass of a photon is zero, and the ADM energy-momentum transforms a lot like the standard 4-vectors in special relativity (see above). We also expect that the ADM energy should be equal to the Komar energy. Given this, we can guess that the ADM momentum gets multiplied by the same "red-shift" factor as the Komar energy.

Last edited: Sep 11, 2006
3. Sep 11, 2006

### MeJennifer

The photon does not change direction at all, it just appears so from our frame of reference. In reality the path of the photon is a geodesic and thus it simply follows the curvature of space.

4. Sep 11, 2006

### pervect

Staff Emeritus
While a photon is too small to cause an effect that can actually be observed, note that suns do wobble when large planets orbit them. In fact, this has acatually been observed. This is true in both GR and in Newtonian gravity.

Photons also have momentum, just like planets do. However, the amount is usually small enough that it can be ignored in practice. In theory, though, there is no difference between the wobble caused by an orbiting planet and the wobble caused by orbiting "photons" - it's just that the amount of momentum in the photons is very small.

Last edited: Sep 11, 2006
5. Sep 12, 2006

### exponent137

If we look from distance in flat space, light-ray change direction for 1.75''.
1. I suppose that frequency of this photon is not changed in -+ infinity.
2. We looked sun from the distance (from flat space). I suppose that this distant momentum change equals momentum change of photon.
3. I suppose that momentum change of photon is the same as 1.75'' momentum change?
4. It is very unrealistically that this photon does not cause momentum change.
5. Can mass of sun be defined from large distance so that we ignore general relativity: If we observe sun from distance we see some velocity (because of photon impact). We know momentum change of photon (mcp). Then
mcp=m(sun) * v(sun)?

Last edited: Sep 12, 2006
6. Sep 12, 2006

### Meir Achuz

I agree with that.

7. Sep 12, 2006

### pervect

Staff Emeritus
I think I mostly agree too, with the provision that the frequency of the photon does vary as the photon falls towards the sun. This is where the redshift factor I mentioned comes into play. If one only considers the photon "at infinity", one can neglect this. If one wants to compute the momentum of the system when the photon is at some position other than infinity, one needs to take into account the redshift factor. Essentially I was arguing earlier that one needs to use the energy of the photon "at infinity" to compute its momentum when the photon is not at infinty, i.e. close to the sun, rather than the locally measured energy of the photon . The locally measured energy of the photon will be higher than the energy-at-infinty, due to gravitational blueshift. It's probably simpler and safer to avoid considering this issue, though, and only treat the case where the photon is at infinity.

In order to sucessfully assign a momentum to the sun, one still needs asymptotically flat space-time, or a close enough approximation of the same. One will automatically have this if one idealizes the problem so that the photon and the sun are alone in an infinite vacuum space-time.

8. Sep 12, 2006

### MeJennifer

Yes from any other frame of reference than the photon's one.

However I consider it incorrect to assert that gravity influences the frequency of the photon itself. And neither does gravity influences it's momentum.
Space-time curvature and gravitational "pull" does not influence frequencies in the local frame, and neither does it influence the object's momentum. The observed effects from other frames of references are relativistic.

Last edited: Sep 13, 2006
9. Sep 13, 2006

### exponent137

I thougth that frequency of photon is equal from -infinity to +infinity. In mid time path of photon is curved by sun.

10. Sep 13, 2006

### pervect

Staff Emeritus
It depends on how you measure it, exactly. If you use local clocks, the frequency of the photon changes as the photon gets closer to the sun. This is usually called "Gravitational redshift", and was demonstrated in the Harvard tower experiment. Note that in this case, the locally measured frequency of the photon would increase as it neared the sun, so it would be blueshift rather than redshift.

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/gratim.html

If you use a coordinate based approach, rather than the simpler local approach, in the Schwarzschld coordinate system the energy-at-infinty E0 is constant for the photon during its orbit. Note, however, that E0 varies during the orbit (E0 = g00 E0) and that E = sqrt(|E0 E0|) also varies. In particular, we know that g00 = 1/(1-2GM/rc^2) for the Schwarzschild coordinate system.

[add] I've been talking about energies here, but E = h v, so the two are proportional. The fact that clocks run at different rates causes a large amount of the complexity here - one of the questions here is "whose clock do you use to measure the frequency of the photon". A more subtle, but equally important, question is how does one measure the frequency of a photon when that photon is not at the same location as the observer. This gets into issues of parallel transport, and of the relativity of simultaneity. The easiest solution is to avoid the difficulty by always measuring the frequency of the photon with the clock of a local observer. When this approach is adopted, the frequency of the photon does vary as it gets closer to the sun.

Last edited: Sep 13, 2006
11. Sep 13, 2006

### WhyIsItSo

I read the Wiki you pointed to earlier. All the formulas for momentum appear (to my limited understanding) to contain a unit for mass.

Since a photon is massless, I for one am none the wiser as to where a photon's momentum is coming from. In fact, I would have thought MeJenifer was correct is stating the photon doesn't change course at all, rather is merely following the "straight" path of local space-time.

What am I missing?

12. Sep 13, 2006

Staff Emeritus
A photon's momentum is given by Einstein's other 1905 equation: $$\mathbf{p} = h\nu\mathbf{n}$$. where $$\nu$$ is the photon's frequency, h is Planck's constant, and $$\mathbf{n}$$ is a unit vector in the direction of the photon's motion. As you see it doesn't require a mass.

13. Sep 13, 2006

### NateTG

In classical physics, momentum is indeed tied to rest mass. However in relativity (both special and general) momentum, mass, and such are a bit different.

14. Sep 13, 2006

### WhyIsItSo

Hmm, does that mean momentum does not imply inertia?

selfAdjoint, does that equation mean light would be defracted by passing near the sun? Just like passing through some medium?

15. Sep 13, 2006

### Staff: Mentor

In classical electromagnetism, electromagnetic fields carry momentum as well as energy.

16. Sep 13, 2006

### pervect

Staff Emeritus
The idea that it is necesary to have mass in order to have momentum is an (incorrect) carry-over from Newtonian mechanics.

Momeuntum in relativity is not equal to mass*velocity. It's easy enough to learn the new defintions if one does not "cling" to outdated notions inherited from Newtonian mechanics.

17. Sep 14, 2006

### WhyIsItSo

This is an issue about which I am suffering persistent confusion.

In terms of relativity, can you provide a verbal definition of momentum.

I'm stumbling over my mental "tie" between momentum and mass. I also thought inertia and momentum were intimately related (if not actually the same thing). Very, very confused here.

18. Sep 16, 2006

### pervect

Staff Emeritus
Probably the most fundamental defintion of momentum is that due to Noether's theorem.

http://en.wikipedia.org/wiki/Noether's_theorem

The important thing about momentum is that it is a conserved quantity. Noether's theorem guarantees that any physical system (one based on the sort of physics we use, the sort of physics that can be derived from an action principle) that has space translation symmetry will have a conserved quantity.

Note that Newtonian physics, special relativity, AND general relativity can all be expressed in terms of an action principle, if desired, so therefore Noether's theorem applies to all of them. (It is more difficult to find the necessary symmetries in GR, however, even though it does satisfy the "action principle" requirement).

We can also see via Noether's theorem why energy and momentum are so closely related, something that we cant see easily in any other manner.

Momentum is just a consequence of space translation invariance, and energy is just a consequence of time translation invariance. Thus momentum and energy "mix" together in relativity the same way that time and space do.

I'm not sure if you are familiar with how time and space "mix", the classic illustration of this can be found in Taylor & Wheeler's space-time physics. Basically, space and time "mix" together to from an invariant, the Lorentz interval, in very much the way that north-south and east-west distances "mix" together to from an invariant which we simply call "distance". The major (and important) difference is one of sign. The Euclidean distance formula is

ds^2 = dx^2 + dy^2

The Lorentz formula is

ds^2 = dx^2 - dt^2

Anyway, Noether's theorem is the abstract defintion that most clearly defines momentum. It's possible to approach the material in a somewhat less abstract manner, which is more understandable, perhaps, but less fundamental. The approach one takes here is that "momentum must be a conserved quantity".

One can then show that in Newtonian mechanics mv is such a conserved quantity, and call it momentum.

In SR, one can then show that mv does not give rise to a conserved quantity, but that mv / sqrt(1-(v/c)^2) does give rise to a conserved quantity. Without Noether's theorem this seems almost "accidental", with Noether's theorem one can give a "reason" why momentum is conserved.

Now the fact that mv / sqrt(1-(v/c)^2) doesn't , unfortunately, directly address the momentum of light. One can see that m is zero, and so is sqrt(1-(v/c)^2), so that we have 0/0. The correct formula for the momentum of light turns out to be p = E/c.

Of course you've probably already seen these equations. The basic difficulty is that you just seem to refuse to accept them. Basically you seem to have learned Newtonian mechanics "too well", and can't do the unlearning of Newtonian mechanics needed to learn relativity. It's really not hard - you just have to decide to not belive that p = mv anymore, and replace it with the relativistic versions: p = E/c for light, and p=mv/sqrt(1-(v/c)^2) for particles with a rest mass. Note that both of these equations satisfy the more fundanmental relationship

E^2 - (pc)^ = (m c^2)^2

This fundamental relationship says that (mc^2) is a Lorentz invariant of the energy-momentum 4-vector. It can bea viewed as a direct consequence of Noether's theorem and the invariance of the Lorentz interval, ds^2 = dx^2 + dy^2 + dz^2 - dt^2

19. Sep 22, 2006

### exponent137

Gravitational force of photon on the sun

I continue with this problem. (I believe that photon act on the sun with force)

A photon, which flies far distant from the sun, act on the sun with force F=k/r^2. k is proportionality factor and r is distance from the sun. Change of photon's (sun's) momentum is proportional to (sin(theta))^2, and theta is proportional to 1/r. But, if we has current of photons extended to infinity, uniformly dense and it flies in one direction, it act on the sun with infinite force. (Calculation show that this force is logarithmically divergent with r. F=integral of (k dr/r))

But, this is doubtful. :surprised Probabably something special which make calculation converent should be respected in calculation? I used some corrections to this approximation, but all are divergent.

Last edited: Sep 22, 2006
20. Sep 22, 2006

### pervect

Staff Emeritus
The problems with this Newtonian analysis should become apparent when one remembers that light deflects differently (twice as much) in GR than it does with the Newtonian analysis.

You might get a correct order of magitude estimate, but don't be surprised if the detailed answers are off by a factor of as much as 2:1 with this sort of approach.

Basically, if you want to do GR, you have to learn it first.